Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/ Semiconductor Device Modeling and Characterization – EE5342 Lecture 4 – Spring 2011 Professor Ronald L. Carter ronc@uta.edu http://www.uta.edu/ronc/

First Assignment e-mail to listserv@listserv.uta.edu In the body of the message include subscribe EE5342 This will subscribe you to the EE5342 list. Will receive all EE5342 messages If you have any questions, send to ronc@uta.edu, with EE5342 in subject line. ©rlc L04 26Jan2011

Second Assignment Submit a signed copy of the document that is posted at www.uta.edu/ee/COE%20Ethics%20Statement%20Fall%2007.pdf ©rlc L04 26Jan2011

Semiconductor Electronics - concepts thus far Conduction and Valence states due to symmetry of lattice “Free-elec.” dynamics near band edge Band Gap direct or indirect effective mass in curvature Thermal carrier generation Chemical carrier gen (donors/accept) ©rlc L04 26Jan2011

Counting carriers - quantum density of states function 1 dim electron wave #s range for n+1 “atoms” is 2p/L < k < 2p/a where a is “interatomic” distance and L = na is the length of the assembly (k = 2p/l) Shorter ls, would “oversample” if n increases by 1, dp is h/L Extn 3D: E = p2/2m = h2k2/2m so a vol of p-space of 4pp2dp has h3/LxLyLz ©rlc L04 26Jan2011

QM density of states (cont.) So density of states, gc(E) is (Vol in p-sp)/(Vol per state*V) = 4pp2dp/[(h3/LxLyLz)*V] Noting that p2 = 2mE, this becomes gc(E) = {4p(2mn*)3/2/h3}(E-Ec)1/2 and E - Ec = h2k2/2mn* Similar for the hole states where Ev - E = h2k2/2mp* ©rlc L04 26Jan2011

Fermi-Dirac distribution fctn The probability of an electron having an energy, E, is given by the F-D distr fF(E) = {1+exp[(E-EF)/kT]}-1 Note: fF (EF) = 1/2 EF is the equilibrium energy of the system The sum of the hole probability and the electron probability is 1 ©rlc L04 26Jan2011

Fermi-Dirac DF (continued) So the probability of a hole having energy E is 1 - fF(E) At T = 0 K, fF (E) becomes a step function and 0 probability of E > EF At T >> 0 K, there is a finite probability of E >> EF ©rlc L04 26Jan2011

Maxwell-Boltzman Approximation fF(E) = {1+exp[(E-EF)/kT]}-1 For E - EF > 3 kT, the exp > 20, so within a 5% error, fF(E) ~ exp[-(E-EF)/kT] This is the MB distribution function MB used when E-EF>75 meV (T=300K) For electrons when Ec - EF > 75 meV and for holes when EF - Ev > 75 meV ©rlc L04 26Jan2011

Electron Conc. in the MB approx. Assuming the MB approx., the equilibrium electron concentration is ©rlc L04 26Jan2011

Electron and Hole Conc in MB approx Similarly, the equilibrium hole concentration is po = Nv exp[-(EF-Ev)/kT] So that nopo = NcNv exp[-Eg/kT] ni2 = nopo, Nc,v = 2{2pm*n,pkT/h2}3/2 Nc = 2.8E19/cm3, Nv = 1.04E19/cm3 and ni = 1E10/cm3 ©rlc L04 26Jan2011

Calculating the equilibrium no The ideal is to calculate the equilibrium electron concentration no for the FD distribution, where fF(E) = {1+exp[(E-EF)/kT]}-1 gc(E) = [4p(2mn*)3/2(E-Ec)1/2]/h3 ©rlc L04 26Jan2011

Equilibrium con- centration for no Earlier quoted the MB approximation no = Nc exp[-(Ec - EF)/kT],(=Nc exp hF) The exact solution is no = 2NcF1/2(hF)/p1/2 Where F1/2(hF) is the Fermi integral of order 1/2, and hF = (EF - Ec)/kT Error in no, e, is smaller than for the DF: e = 31%, 12%, 5% for -hF = 0, 1, 2 ©rlc L04 26Jan2011

Equilibrium con- centration for po Earlier quoted the MB approximation po = Nv exp[-(EF - Ev)/kT],(=Nv exp h’F) The exact solution is po = 2NvF1/2(h’F)/p1/2 Note: F1/2(0) = 0.678, (p1/2/2) = 0.886 Where F1/2(h’F) is the Fermi integral of order 1/2, and h’F = (Ev - EF)/kT Errors are the same as for po ©rlc L04 26Jan2011

Degenerate and nondegenerate cases Bohr-like doping model assumes no interaction between dopant sites If adjacent dopant atoms are within 2 Bohr radii, then orbits overlap This happens when Nd ~ Nc (EF ~ Ec), or when Na ~ Nv (EF ~ Ev) The degenerate semiconductor is defined by EF ~/> Ec or EF ~/< Ev ©rlc L04 26Jan2011

Donor ionization The density of elec trapped at donors is nd = Nd/{1+[exp((Ed-EF)/kT)/2]} Similar to FD DF except for factor of 2 due to degeneracy (4 for holes) Furthermore nd = Nd - Nd+, also For a shallow donor, can have Ed-EF >> kT AND Ec-EF >> kT: Typically EF-Ed ~ 2kT ©rlc L04 26Jan2011

Donor ionization (continued) Further, if Ed - EF > 2kT, then nd ~ 2Nd exp[-(Ed-EF)/kT], e < 5% If the above is true, Ec - EF > 4kT, so no ~ Nc exp[-(Ec-EF)/kT], e < 2% Consequently the fraction of un-ionized donors is nd/no = 2Nd exp[(Ec-Ed)/kT]/Nc = 0.4% for Nd(P) = 1e16/cm3 ©rlc L04 26Jan2011

Classes of semiconductors Intrinsic: no = po = ni, since Na&Nd << ni =[NcNvexp(Eg/kT)]1/2,(not easy to get) n-type: no > po, since Nd > Na p-type: no < po, since Nd < Na Compensated: no=po=ni, w/ Na- = Nd+ > 0 Note: n-type and p-type are usually partially compensated since there are usually some opposite- type dopants ©rlc L04 26Jan2011

References *Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989. **Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago. M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003. 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. 2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981. ©rlc L04 26Jan2011